I have an assignment requiring me to make 1-100 using 2,0,2,2 and I have done the first 32.
You can use +-x/ square root decimal point power factorial double factorial () permutation and combination.
The deadline of the assignment is 7th May 15:59:59 UTC
I am stuck,can anybody help me please? Thank you.
1=2x0+2/2
2=-20+22
3=2x0!/2+2
4=2x0+2+2
5=2-0+2+2
6=2+0+2+2
7=2+0!+2+2
8=2x2x2+0
9=2x2x2+0!
10=sqrt(20/2)^2
11=(2+2)!/2-0!
12=(2+2)!/2-0
13=(2+2)!/2+0!
14=(2+2)!!+2+0!
15=2^(2x2)-0!
16=2^(2+2)+0
17=2^(2+2)+0!
18=(2+2)!-(2+0!)!
19=22-2-0!
20=-2-0+22
21=-2+0!+22
22=2x0+22
23=(2+2)!-2+0!
24=2-0+22
25=(2+2)!+2-0!
26=(2+2)!+2+0
28=22+(2+0!)!
29=(.2)^(-0!)+(2+2)!
30=(2+2)!+(2+0!)!
31=(.2)^(-2)+(2+0!)!
32=2^(2+2+0!)
Here's a few ....maybe I can think of some others
[.2^-2 ] *2 - 0! = 49
[.2^-2] * 2 + 0! = 51
[.2^-2] * 2 / 0! = 50
C (20, 2) / 2 = 95
(2 + 2)! * ( 2! + 0!) = 72
(2 + 2)! + 20 = 44
(.2^-2) + 20 = 45
[ 2^(2! + 0!) ] ^2 = 64
20 + 22 = 42
(20 - 2 ] * 2 = 36
[ 22 * 2 ] - 0! = 43
[ 2 (20) ] - 2 = 38
[ .2^--2 + 0! ] * 2 = 52
[.2^-2 - 0! ] * 2 = 48
(20/2)^2 = 100
No program....just off the top of my head.....
I did something like this before.....I don't believe that either me, the questioner (or his teacher) could find all of them....I think we found something like 83 of them
I'll see if I can find the post....
Tools
5=Sqrt((.2)^(-2))=(.2)^(-0!)
48=((2+0!)!)!!
8=(2+2)!!
100=(.(0!))^(-2)
6=(2+0!)!
Tools
5=Sqrt((.2)^(-2))=(.2)^(-0!)
48=((2+0!)!)!!
8=(2+2)!!
100=(.(0!))^(-2)
6=(2+0!)!
15=5!!
Allowing for the left factorial
! [ 2 * 2 ] ^2 + 0! = ! [ 4]^2 + 1 = [9]^2 + 1 = 82
! [ 2 + 2 + 0! ] * 2 = ! [ 5 ] *2 = [44] * 2 = 88
([ 2 + 0! ] !) !! + 20 = [ [ 3! ]!! + 20 = 6!! + 20 = 48 + 20 = 68
C ( !(2 + 2) , 2 + 0!) = C ( 9,3) = 84